Editing Floyd–Warshall algorithm (section)

==Time complexity==
Let <math>n</math> be <math>|V|</math>, the number of vertices. To find all <math>n^2</math> of 
<math>\mathrm{shortestPath}(i,j,k)</math> (for all <math>i</math> and <math>j</math>) from those of
<math>\mathrm{shortestPath}(i,j,k-1)</math> requires [[big theta|<math>\Theta(n^2)</math>]] operations. Since we begin with
<math>\mathrm{shortestPath}(i,j,0) = \mathrm{edgeCost}(i,j)</math> and compute the sequence of <math>n</math> matrices <math>\mathrm{shortestPath}(i,j,1)</math>, <math>\mathrm{shortestPath}(i,j,2)</math>, <math>\ldots</math>, <math>\mathrm{shortestPath}(i,j,n)</math>, each having a cost of <math>\Theta(n^2)</math>,
the total [[time complexity]] of the algorithm is <math>n \cdot \Theta(n^2) = \Theta(n^3)</math>.<ref name=":1" /><ref>{{cite book
| last1 = Baras 
| first1 = John 
| last2 = Theodorakopoulos 
| first2 = George 
| date = 2022 
| title = Path Problems in Networks 
| publisher = Springer International Publishing
| isbn = 9783031799839
}}</ref>